troubleshooting Question

Oriented bounding rectangle for a line

Avatar of Bill-Hanson
Bill-HansonFlag for United States of America asked on
Game ProgrammingMath / ScienceAlgorithms
6 Comments1 Solution486 ViewsLast Modified:
I'm working on a flowchart application, and need to define the click-able region for a shape connector (line-segment).

To do this, I a need to generate the vertexes for an oriented rectangle that fully encloses an arbitrary line.  In addition, I need to be able to specify some padding (in pixels) such that the resulting rectangle will be slightly larger than the line.

The code below is the closest solution I have found, but the resulting rectangle does not include the padding at the end of the line, only at the sides.

The image below illustrates the difference between what the GetSegmentBounds function does and what I need it to do.

For reference, I found the code below here:  https://www.experts-exchange.com/Q_22951696.html#20261839
protected Point[] GetSegmentBounds(Point p1, Point p2, Double offset)
        {
            Point[] rect = new Point[4];
            Double len = Math.Sqrt((Double)((p1.X - p2.X) * (p1.X - p2.X) + (p1.Y - p2.Y) * (p1.Y - p2.Y)));// +(offset * 2);
            rect[0].X = (Int32)(p1.X - (p1.Y - p2.Y) * offset / len);
            rect[0].Y = (Int32)(p1.Y + (p1.X - p2.X) * offset / len);
            rect[1].X = (Int32)(p1.X + (p1.Y - p2.Y) * offset / len);
            rect[1].Y = (Int32)(p1.Y - (p1.X - p2.X) * offset / len);
            rect[2].X = (Int32)(p2.X + (p1.Y - p2.Y) * offset / len);
            rect[2].Y = (Int32)(p2.Y - (p1.X - p2.X) * offset / len);
            rect[3].X = (Int32)(p2.X - (p1.Y - p2.Y) * offset / len);
            rect[3].Y = (Int32)(p2.Y + (p1.X - p2.X) * offset / len);
            return rect;
        }
Rectangles.jpg
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